Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
long long x[10005];
long long y[10005];
vector<pair<int, int> > xorr1;
vector<pair<int, int> > xorr2;
vector<int> place(30, -1);
vector<bool> was(10005, false);
vector<int> need(10005, -1);
void initialization() {
  cin >> n;
  for (long long i = 0; i < n; ++i) cin >> x[i];
  for (long long i = 0; i < n; ++i) cin >> y[i];
}
void solve() {}
void print() {}
void flush() {}
void swapp(int i, int j, long long* x, vector<pair<int, int> >& xorr) {
  if (i == j) return;
  x[i] ^= x[j];
  xorr.push_back(make_pair(i, j));
  x[j] ^= x[i];
  xorr.push_back(make_pair(j, i));
  x[i] ^= x[j];
  xorr.push_back(make_pair(i, j));
}
void get_basis(long long* x, vector<pair<int, int> >& xorr, bool q) {
  int cur = 0;
  vector<bool> used(10005, false);
  int j = 29;
  while (j >= 0) {
    int i = cur;
    while (i < n && (used[i] || !((x[i] >> j) & 1))) ++i;
    if (i == n) {
      --j;
      continue;
    }
    used[cur] = true;
    if (i != cur) swapp(i, cur, x, xorr);
    if (q) place[j] = cur;
    for (int k = 0; k < n; ++k) {
      if (cur != k) {
        if ((x[k] >> j) & 1) {
          x[k] ^= x[cur];
          xorr.push_back(make_pair(k, cur));
        }
      }
    }
    ++cur;
    --j;
  }
}
int main() {
  ios::sync_with_stdio(0);
  initialization();
  solve();
  print();
  get_basis(x, xorr1, 1);
  get_basis(y, xorr2, 0);
  for (int i = 0; i < n; ++i) {
    if (y[i] == 0) continue;
    int cur = 0;
    vector<pair<int, int> > xorr;
    bool q = false;
    int p;
    for (int j = 29; j >= 0; --j) {
      if (((y[i] >> j) & 1) && place[j] >= 0) {
        if (!q) p = place[j];
        q = true;
        cur ^= x[place[j]];
        if (p != place[j]) xorr.push_back(make_pair(p, place[j]));
      }
    }
    if (!q || cur != y[i]) {
      cout << -1;
      return 0;
    }
    was[p] = true;
    need[i] = p;
    x[p] = y[i];
    for (int j = 0; j < xorr.size(); ++j) xorr1.push_back(xorr[j]);
  }
  for (int i = 0; i < n; ++i) {
    if (!was[i]) {
      x[i] ^= x[i];
      xorr1.push_back(make_pair(i, i));
    }
  }
  for (int i = 0; i < n; ++i) {
    if (need[i] >= 0) swapp(i, need[i], x, xorr1);
  }
  reverse(xorr2.begin(), xorr2.end());
  cout << xorr1.size() + xorr2.size() << endl;
  for (int i = 0; i < xorr1.size(); ++i)
    cout << xorr1[i].first + 1 << ' ' << xorr1[i].second + 1 << endl;
  for (int i = 0; i < xorr2.size(); ++i)
    cout << xorr2[i].first + 1 << ' ' << xorr2[i].second + 1 << endl;
  return 0;
}